Solution/Correction standard, second Test Mathematics A + B1; November 7, 2014.

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1 Solution/Correction standard, second Test Mathematics A + B1; November 7, 014. Kenmerk : Leibniz/toetsen/Re-Exam-Math-A-B1-141-Solutions Course : Mathematics A + B1 (Leibniz) Vakcode : Date : November 7, 014 Solution/Correction standard 1. [4 pt] A 4 A [ 0, 1 4 ) (, 1 ] A 4 A [ 1, 1 4 ) A 4 A [ 1, 1 ] 10 k A k [ 0, 1 10 ) (, 1 ] False boundary (e.g. ( 1, 1) 4 instead of [ 1, 1 )): 0. pt. 4. [ pt] The statement is false (truth value 0) if U Z. [0. pt] E.g. take x 0 and y 1: then there does not exist any z Z with x < z < y. [0. pt] The statement is true (truth value 1) if U R. [0. pt] E.g. take, for any x, y R, z x+y (the average of x and y). Then x < z < y. [0. pt]. (a) [ pt] (i) Proof of A B A B. Let x A B. Then x A B. So x A or x B. Hence x A or x B. And so x A B. (ii) Proof of A B A B. Let x A B. Then x A or x B. So x A or x B. Hence x A B. And so x A B. In the proof of part (ii) it suffices to note that it is the converse of the proof of part (i); or one can prove both directions at once using "if and only if" connectives. Important is that it is clear that both directions are proved. (b) [ pt] Base step for n 1: 1 i(i + 1) 1 and also 1 (1 + 1) (1 + ) 6. 1

2 So the statement is correct for n 1. Induction step: [0. pt] Let k 1 and suppose that: k k(k + 1)(k + ) i(i + 1). (Induction hypothesis: IH). k+1 (k + 1)(k + )(k + ) We must show that IH implies: i(i + 1). [0. pt] k+1 k Well: i(i + 1) i(i + 1) + (k + 1)(k + ). By IH this expression is equal to k(k + 1)(k + ) Now it remains to show that k(k + 1)(k + ) + (k + 1)(k + ) This is straightforward: k(k + 1)(k + ) + (k + 1)(k + ) (k + 1)(k + ) + (k + 1)(k + ). (k + 1)(k + )(k + ). ( ) k + 1 [0. pt] (k + 1)(k + )(k + ). [0. pt] Now we obtain from the principle of mathematical induction that for all n 1: n n(n + 1)(n + ) i(i + 1). (From the proof it must be crystal clear whàt is supposed and whàt must be proved. In case of nonsense formulations like Suppose it is correct FOR ALL n, so it also holds for n + 1 : at most 1 pt for the entire exercise) 4. [ pt] Three possibilities: (6 books of A, of B and of C), ( books of A, 6 of B and of C) and ( books of A, of B and ( ) 6 of C). [0. pt] 8 In the first possibility, there are ways to choose 6 books of author A and ( ) 6 ( ) 10 1 ways to choose books of author B and ways to choose books of author C. [0. pt] So, by the rule of product, the first possibility can be performed in

3 ( ) 8 6 ( ) 10 ( ) 1 ways. ( )( 8 10 Similarly, the second possibility can be performed in ( )( )( ) the third possibility can be performed in ways. 6 )( 1 ) [0. pt] ways, and [0. pt] Therefore, by the sum rule, the number of different selections of 16 books equals: ( )( )( ) ( )( )( ) ( )( )( ) [ pt] The differential equation is linear, it is of the form y +P (x)y Q(x) with P (x) 1 x and Q(x) The integrating factor is x v(x) e P (x) dx e 1/x dx e ln x x. Multiplying the equation y + P (x)y Q(x) with v(x) x yields the original equation. Remark: If a student notices that xy + y d ( ) x y(x) directly: dx The general solution is v(x)y (x) + 1 ( x v(x)y(x) v(x) ) x ( v(x)y (x) + v (x)y(x) x ) x d ( ) v(x)y(x) x + 1 dx d ( ) x y(x) x + 1 dx x y(x) 1 x + x + C, hence y(x) 1 x C x. Use the initial condition to find C: The solution therefore is: y(1) C c 1. y(x) 1 x x (x + 1). x

4 6. (a) Multiply numerator and denominator of z with 1 + i: ( 1 ) i 1 i ( ) i 1 + i 1 i 1 + i ( ) [ (1 ) ( ) ] i + i i. (b) For the modulus, calculate z : z Re z + Im z ( ) + 1 4, hence z. Alternative 1: For the argument θ we have [0. pt] tan θ Im z Re z 1 1. This means that θ π 6 or θ 7π 6 that θ π 6. Alternative : Use a picture, see Figure 1.. From Re z > 0 we readily conclude [0. pt] [0. pt] i z 0 1 π 6 R Figure 1: z lies on a circle with radius (assignment 6). 4

5 7. [ pt] Let z x + iy with x, y R. Then 1 z x i y x + y x x + y + y x + y i. For 1/z we then have 1 z ( ) ( x + y ) x + y x + y x + y ( x + y ) 1 x + y 1 z. Taking square roots then gives the desired result. 8. (a) [ pt] The correct answer is (D). (b) If a correct argumentation is given: For instance: From the equation one derives that y > 0 in the first quadrant and y < 0 in the fourth quadrant, which eliminates (A) and (B). If y 0 then y 0, which eliminates (C). A wrong sketch ( a curve not following the slope lines; a curve not passing the point (0, 0.) ) : [0 pt] A correct sketch only in picture (D): If a student gives the wrong answer in (a), say (B), then a correct sketch only in picture (B): If a student submits correct sketches in more than one picture (even if (D) belongs to them): [0. pt] See Figure for the correct sketches. 9. [ pt] Step 1 (total: pt): solve the homogeneous equation y + 9y 0. The corresponding auxiliary or characteristic equation is λ This equation has two imaginary roots: λ i and λ i. Therefore the general solution of the homogeneous equation is y(x) c 1 cos(x) + c sin(x).

6 (A) (B) (C) (D) Figure : all possible solutions through (0, 0.) (assignment 8) Remark: Another possibility is to give the complex solution y(x) α 1 e ix + α e ix. [0. pt] The other [0. pt] is earned by having the final answer real. Step (total: 1 pt): find a particular solution. We try a first order polynomial y p (x) a + bx [0. pt] with unknown constants a and b. Notice that y p (x) a and y p (x) 0, hence y p + 9y p 9a + 9bx 6

7 From this readily follows that a 1 and b. Step (total: pt): determine the constants c 1 and c. The general solution is From y(0) 1 follows [0. pt] y(x) (c 1 cos(x) + c sin(x)) x. (1) 1 y(0) 1 (c c 0) c 1 + 1, hence c 1 0. The general solution (1) simplifies to Differentiate y(x): From y (0) 8 follows y(x) c sin(x) x. y (x) c cos(x) +. 8 y (0) c cos(0) + c +, hence c 8. The solution of the initial value problem is y(x) sin(x) x. 10. (a) [ pt] First calculate P # R» and P # Q:» P R 1,, 1, 1, 1 0, 4, 4 and P Q 1,, 1, 1, 1,,. [0. pt] Calculate the cross product of P # R» and P # Q:» [ ] u P # R» P # Q» , 8, 8. [1. pt] (b) The normal equation of the plane is (x p) u 0, [0. pt] with p a point in the plane. For instance with p OP : gives simplified: ( x, y, z 1, 1, 1 ) 4, 8, 8 0, 4(x 1) + 8(y + 1) + 8(z + 1) 0, x + y + z. [0. pt] 7

8 (c) Notice that RP and RQ are both vectors parallel to the plane of (b), hence RP RQ is perpendicular on this plane. Since u is also perpendicular to the plane, the assertion most hold. The same argument holds for QP QR. Or: Since P,Q, and R lie in the plane calculate in part (b), the vector RP RQ is a normal vector to this plane. The direction of the normal vector is unique for the plane, and so QP QR must be a multiple of u. The same argument holds for QP QR. A more elaborate way is to calculate RP RQ and QP QR directly: [ ] RP RQ , 8, 8, 1 1 hence u RP RQ (in other words: α 1), and QP QR [ ] 1 1 hence u QP QR (in other words: β 1). 4, 8, (a) Solve the equation u v 0. Since u v α, the only solution is α 0. (b) [ pt] Use the identity u v u v cos θ, where θ is the angle between u and v. u v u v cos ( 1 4 π) [0. pt] α 6 α α α + 1. [0. pt] () By squaring the left- and righthand side we get 4α ( α + 1 ) which gives α ±. α, [0. pt] Now notice that from equation () follows that α must be positive, hence the only solution is α. [0. pt] 8

Math 113 (Calculus 2) Exam 4

Math 113 (Calculus 2) Exam 4 Math 3 (Calculus ) Exam 4 November 0 November, 009 Sections 0, 3 7 Name Student ID Section Instructor In some cases a series may be seen to converge or diverge for more than one reason. For such problems